1. Here are a few of the possible solutions.

2. We call the people at the party P0, P1, ..., P8, and M. Pi is the person who shook i hands, and M is the mathematician. Now P8 shook eight hands, and there are only 10 people at the party. So the only people she (or he) did not shake hands with were herself (or himself) and her (or his) spouse. On the other hand P0 did not shake any hands. It follows that P0 and P8 are a couple. Now P7 is not P0's spouse and did not shake hands with P0. Except for P0 and her (or his) spouse, P7 shook hands with everyone else. Now P1 only shook hands with P8 and no one else, so P1 and P7 must be a couple. Continuing in this way, we discover that P2 and P6 are a couple, and so are P3 and P5. So the mathematician's husband is P4, who shook four hands. It also follows that M shook four hands.

3. Chuck has a black hat. The proof of this is a proof by contradiction. Assume first that Chuck has a white hat. Case 1: Beth has a white hat. Case 2: Beth has a black hat. In Case 1, both Chuck and Beth have a white hat. Abe, seeing two white hats, would know that his hat is black. He didn't, and therefore we have a contradition. In Case 2, Chuck has a white hat while Beth has a black hat. Beth hears Abe say that he cannot tell the color of his hat. From this Beth concludes that her own hat could not have been white, because she knows that Chuck has a white hat and if she also had a white hat then Abe would know that his hat was black. Thus Beth can conclude that her hat is black. This is also a contradiction since we know that she could not tell the color of her own hat. Since both cases lead to a contradiction, we can conclude that the assumption must have been false, and thus Chuck must have a black hat.

--Shahriar Shahriari
Associate Dean of the College and Associate Professor of Mathematics